Maltose $+ H _2 O \longleftrightarrow 2$ Glucose
Calculate the equilibrium constant under standard transformed conditions ( $K_{e q}^{\prime}$ ) knowing that for this reaction $\Delta G^{* \prime}=-15.5 kJ . mol ^{-1}$. Make sure to use the correct unit for $K_{e q}^{\prime}$.
A. 0.527 mol
B. $0.527 mol^{-1} L^{-1}$
C. $526.724 L.mol^{-1}$
D. 526.724 (unitless)
E. $526.724 mol^{-1} L^{-1}$
Answer (2)
To find the equilibrium constant K' {eq} for the reaction Maltose + H₂O ↔ 2 Glucose with ΔG' = -15.5 kJ/mol, we calculated K' {eq} as approximately 526.724, which is dimensionless. This was done using the equation K'_{eq} = e^{-ΔG'/(RT)}. Therefore, the correct choice is D. 526.724 (unitless).
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We are given the Gibbs free energy change Δ G ′ = − 15.5 kJ/mol for the reaction Maltose + H 2 O ⟷ 2 Glucose.
We use the formula Δ G ′ = − RT ln K e q ′ to find the equilibrium constant K e q ′ .
Rearranging the formula, we get K e q ′ = e − RT Δ G ′ , where R = 8.314 J/(mol K) and T = 298 K .
Substituting the values, we find K e q ′ ≈ 521.195 , which is dimensionless. The final answer is 521.195
Explanation
Problem Setup and Given Data We are given the standard transformed Gibbs free energy change ( Δ G ′ ) for the reaction:
Maltose + H 2 O ⟷ 2 Glucose
We need to calculate the equilibrium constant under standard transformed conditions ( K e q ′ ) using the relationship:
Δ G ′ = − RT ln K e q ′
where:
Δ G ′ = − 15.5 kJ/mol = − 15500 J/mol
R = 8.314 J/(mol K) is the gas constant
T = 298 K is the standard temperature
We will rearrange the equation to solve for K e q ′ .
Rearranging the Formula and Substitution First, we rearrange the equation to solve for K e q ′ :
ln K e q ′ = − RT Δ G ′
K e q ′ = e − RT Δ G ′
Now, we substitute the given values:
K e q ′ = e − ( 8.314 J/(mol K) ) ( 298 K ) − 15500 J/mol
K e q ′ = e 8.314 × 298 15500
K e q ′ = e 2477.572 15500
K e q ′ = e 6.256
Calculating K'eq and Determining Units Now, we calculate the value of K e q ′ :
K e q ′ = e 6.256 ≈ 521.195
Therefore, the equilibrium constant under standard transformed conditions is approximately 521.195.
To determine the units of K e q ′ , we analyze the reaction:
Maltose + H 2 O ⟷ 2 Glucose
K e q ′ = [ Maltose ] [ H 2 O ] [ Glucose ] 2
If we assume the concentrations are in mol/L, the units of K e q ′ would be:
( mol/L ) ( mol/L ) ( mol/L ) 2 = 1
Thus, K e q ′ is dimensionless (unitless).
Final Answer The equilibrium constant under standard transformed conditions ( K e q ′ ) for the reaction is approximately 521.195, and it is dimensionless.
Examples
Understanding equilibrium constants is crucial in various real-life applications, especially in the food and beverage industry. For example, in the production of high-fructose corn syrup, enzymes are used to convert glucose into fructose to achieve a sweeter product. By knowing the equilibrium constant for the isomerization reaction, manufacturers can optimize the reaction conditions (temperature, enzyme concentration) to maximize the yield of fructose, leading to a more efficient and cost-effective production process. Similarly, in brewing, controlling the enzymatic breakdown of starches into sugars is essential for achieving the desired alcohol content and flavor profile. The equilibrium constants help brewers fine-tune the fermentation process.
